Let the solution curve x = x ( y ) , 0 < y < (π)/(2) , of the...

Given,

${\left(\log_{e}\left(\cos y\right)\right)}^{2}cosydx - \left(1 + 3x{\log}_{e}\left(\cos y\right)\right)sinydy = 0$

$⇒ \dfrac{dx}{dy} + \left(\dfrac{- 3\tan y}{\ln \cos y}\right)x = \dfrac{\sin y}{{\left(In \cos y\right)}^{2} \cdot \cos y}$

Which is a linear differential equation,

So, Integrating factor will be, $IF = {e}^{3 ∫ \dfrac{- \tan y}{\ln \cos y}dy} = {\left(\ln \cos y\right)}^{3}$

So, solution of differential equation is given by,
$x{\left(\ln \cos y\right)}^{3} = ∫ \dfrac{\sin y}{\cos y}\left(\ln \cos y\right)dy$
$⇒ x{\left(\ln \cos y\right)}^{3} = \dfrac{- {\left(\ln \cos y\right)}^{2}}{2} + c$

Now using the given value of $x\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2\ln 2}$
We get, $c = 0$
Hence, $x = \dfrac{- 1}{2\ln \cos y}$

Now finding, $x\left(\dfrac{\pi}{6}\right) = - \dfrac{1}{2\ln \left(\dfrac{\sqrt{3}}{2}\right)} = \dfrac{1}{\ln 4 - \ln 3}$

Now on comparing with $\dfrac{1}{\log_{e}m - \log_{e}n}$ we get, $m = 4, n = 3$

Hence, $mn = 12$